MHB Continuous mapping and fixed point

[ozkan12]

Let $T$ be a continuous mapping of a complete metric space $X$ into itself such that ${T}^{k}$ is a contraction mapping of $X$ for some positive integer $k$. Then $T$ has a unique fixed point in $X$.

Proof:

${T}^{k}$ has a unique fixed point $u$ in $X$ and $u=\lim_{{n}\to{\infty}}\left({T}^{k}\right)^n{x}_{0}$ ${x}_{0}\in X$ arbitrary.

Also $\lim_{{n}\to{\infty}}\left({T}^{k}\right)^nT{x}_{0}=u$. Hence

$u=\lim_{{n}\to{\infty}}({T}^{k})^nT{x}_{0}$=$\lim_{{n}\to{\infty}}T\left({T}^{k}\right)^n{x}_{0}$ (2)
=$T\lim_{{n}\to{\infty}}\left({T}^{k}\right)^n{x}_{0}$
=$Tu$.İn this proof, I didnt understand, How (2) happened ? Please, can you explain ? Thank you for your attention...Best wishes...

 

[Opalg]

Let $T$ be a continuous mapping of a complete metric space $X$ into itself such that ${T}^{k}$ is a contraction mapping of $X$ for some positive integer $k$. Then $T$ has a unique fixed point in $X$.

Proof:

${T}^{k}$ has a unique fixed point $u$ in $X$ and $u=\lim_{{n}\to{\infty}}\left({T}^{k}\right)^n{x}_{0}$ ${x}_{0}\in X$ arbitrary.

Also $\lim_{{n}\to{\infty}}\left({T}^{k}\right)^nT{x}_{0}=u$. Hence

$u=\lim_{{n}\to{\infty}}({T}^{k})^nT{x}_{0}$=$\lim_{{n}\to{\infty}}T\left({T}^{k}\right)^n{x}_{0}$ (2)
=$T\lim_{{n}\to{\infty}}\left({T}^{k}\right)^n{x}_{0}$
=$Tu$.İn this proof, I didnt understand, How (2) happened ? Please, can you explain ? Thank you for your attention...Best wishes...

Since $T^k$ is a contraction map, $\lim_{{n}\to{\infty}}\left({T}^{k}\right)^ny = u$ for any $y\in X$. In particular, this holds for $y=Tx_0$, so that $\lim_{{n}\to{\infty}}\left({T}^{k}\right)^nT{x}_{0}=u$. But powers of $T$ commute with each other, so $\left({T}^{k}\right)^nT = T\left({T}^{k}\right)^n$. Therefore $u = \lim_{{n}\to{\infty}}T\left({T}^{k}\right)^n{x}_{0}$, as claimed in (2).

 

[ozkan12]

Dear professor,

Firstly, Thank you so much...But why powers of T is commute each other ?

 

[Opalg]

Since $T^k$ is a contraction map, $\lim_{{n}\to{\infty}}\left({T}^{k}\right)^ny = u$ for any $y\in X$. In particular, this holds for $y=Tx_0$, so that $\lim_{{n}\to{\infty}}\left({T}^{k}\right)^nT{x}_{0}=u$. But powers of $T$ commute with each other, so $\left({T}^{k}\right)^nT = T\left({T}^{k}\right)^n$. Therefore $u = \lim_{{n}\to{\infty}}T\left({T}^{k}\right)^n{x}_{0}$, as claimed in (2).

But why powers of T is commute each other ?

Index laws: $\left({T}^{k}\right)^nT = T^{kn+1} = t^{1+kn} = T\left({T}^{k}\right)^n$.

 

[ozkan12]

Dear professor,

İn there $({T}^{k})$ is composite function, isn't it ? That is, İn my opinion, we don't get exponentiate

 

[Ackbach]

In this context, $T^k y=\underbrace{TTT\dots T}_{k \, \text{times}}y,$ so that $T^k$ is shorthand for "compose $T$ with itself $k$ times". So, you are correct.

 

[Opalg]

İn there $({T}^{k})$ is composite function, isn't it ? That is, İn my opinion, we don't get exponentiate

The operation of composition is associative, and therefore obeys the index laws.